Finding the center and radius of a circle given a general degree 2 equation I am trying to find the center and radius of the circle with equation $x^2 + y^2 -6x + 10y + 9 = 0$
 A: Hint:
Rearrange the equation as follows: $$ x^2-6x + y^2 +10y =-9$$ and complete the square in $x$ and $y$
.  You should get something of the form $$(x-a)^2+(y-b)^2 = c^2$$ 
from which you can get the center and the radius.  
All the best.
A: You know the equation of a circle has the form
$$
r^2=(x-a)^2+(y-b)^2.
$$
If you have the equation in this form, you can read off the center and radius.
But you don't have that. You need to do a bit of work to get what you have into the form you want.
The tool to use here is:
Completing the Square 
To complete the square for the expression $ax^2+bx+c$ means to find constants $h$ and $k$ such that   $$
     ax^2+bx+c=a(x-h)^2+k.
    $$  To do this, factor out $a$ and add and subtract $b^2\over4a^2$ in the other factor. 

For your equation
 $$\color{maroon}{x^2} +\color{darkgreen}{ y^2} \color{maroon}{-6x }\color{darkgreen}{+ 10y }\color{maroon}{+ 9}=0$$
you want to complete the square on the $\color{maroon}x$ and on the $\color{darkgreen}y$ terms on the left hand side.
For the $y$ terms:
$$
\eqalign{ y^2+10y  &= y^2+10y +{100\over4}-{100\over4} \cr 
                   &=  y^2+10y +25-25\cr
                   &= (y+5)^2-25 
}
$$
The other part of your equation is easier to work out (note its easy when you include the 9 with it):
$$
x^2-6x+9=(x-3)^2
$$
So
$$\eqalign{
& \color{maroon}{x^2} +\color{darkgreen}{ y^2} \color{maroon}{-6x }\color{darkgreen}{+ 10y }\color{maroon}{+ 9}\cr& \iff
\color{maroon}{(x-3)^2}+ \color{darkgreen}{ (y+5)^2-25} =0\cr &\iff (x-3)^2+  (y+5)^2=25.
}
$$
So the center and radius are...
A: A mechanical way is to start with the usual equation of a circle centered at the point $(a,b)$ and radius $r$ 
$$
(x-a)^{2}+(y-b)^{2}=r^{2},\tag{1}
$$
write it in the form
$$
(x-a)^{2}+(y-b)^{2}-r^{2}=0,
$$
and expand the LHS
$$
\begin{eqnarray*}
(x-a)^{2}+(y-b)^{2}-r^{2} &=&\left( x^{2}-2ax+a^{2}\right) +\left(
y^{2}-2by+b^{2}\right) -r^{2} \\
&=&x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2},
\end{eqnarray*}
$$
so that we get the equivalent equation 
$$
x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2}=0.\tag{2}
$$
Equating the coefficients of $(2)$ to the ones of the given equation
$$
x^{2}+y^{2}-6x+10y+9=0\tag{3}
$$
results in the system of equations 
$$
\begin{eqnarray*}
-2a &=&-6 \\
-2b &=&10 \\
a^{2}+b^{2}-r^{2} &=&9,
\end{eqnarray*}
$$
which is equivalent to
$$
\begin{eqnarray*}
a &=&3 \\
b &=&-5 \\
9+25-r^{2} &=&9.
\end{eqnarray*}
$$
So the center is the point $(a,b)=(3,-5)$ and the radius is $r=5$, because it cannot be negative (the other solution of $9+25-r^{2} =9$ is $-5$).
